T 
369 

73 


UC-NRLF 


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PERSPECTIVE 


TAYLOR 


LIBRARY  ~,_ 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 

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OF 

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NET  BOOK— This  Book  is  supplied 
to  the  trade  on  terms  which  do  not 
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THE  MYRON  C.  CLARK  PUBLISHING  CO. 


CHICAGO 

THE  MYRON  C.  CLARK  PUBLISHING  COMPANY 
1910 


BACKBONE 

OF 

PERSPECTIVE 


By  T.  U.  TAYLOR 

C.  E.,  University  of  Virginia;  M.  C.  E., 

Cornell  University ;  Professor  of  Civil 

Engineering,  University  of  Texas ; 

Member  Am.  Soc.  C.  E. 


CHICAGO 

THE  MYRON  C.  CLARK  PUBLISHING  COMPANY 
1910 


TJ6? 

T3 


Copyright,  1910, 

By 
THE  MYRON  C.  CLARK  PUBLISHING  Co. 


PREFACE 

These  notes  have  been  given  in  the  form  of  lec- 
tures and  drawing-board  exercises  for  many  years. 
They  are  here  reduced  to  print  to  save  time  in  note- 
taking  on  the  part  of  the  student.  A  reader  that 
finds  no  errors  in  these  pages  should  read  again. 
I  am  indebted  to  E.  E.  Howard,  a  former  instructor, 
for  much  valuable  assistance.  Prof.  E.  C.  H. 
Bantel,  and  Instructor  S.  P.  Finch  have  rendered 
substantial  aid.  The  chapter  on  Axometric  Projec- 
tions is  a  modification  of  notes  taken  under  Dr.  W. 
M.  Thornton  of  the  University  of  Virginia,  when  I 
was  a  student  there. 

T.  U.  TAYLOR. 

Austin,  Texas. 

June,  1910. 


212042 


(   UNIVERSITY   ) 

>»IUFORjfife»X 


BACKBONE  OF  PERSPECTIVE 


CHAPTER  ONE. 


Primary  Methods. 

i.  Plan  and  Projection. — When  a  perpendicu- 
lar is  dropped  from  any  point  P  to  a  plane,  the  point 
of  intersection  p  of  the  perpendicular  and  the  plane 
is  called  the  foot  of  the  perpendicular  and  p  is  called 
the  projection  of  P.  If  from  two  points,  A  and  B, 
perpendiculars  are  dropped  on  any  plane,  and  their 
feet  a  and  b  be  joined,  the  line  ab  is  called  the  pro- 
jection of  AB  on  the  given  plane.  The  projection 
of  a  point  on  a  horizontal  plane  is  called  the  plan  of 


Figure  1. 


the  point,  and  its  projection  on  a  vertical  plane  is 
called  its  elevation.    We  shall  denote  the  horizontal 
plane  by  H  in  these  pages. 
2.    Perspective. — The  perspective  of  a  point  P 


2  BACKBONE   OF  PERSPECTIVE. 

( Fig.  i )  with  reference  to  any  other  point  E  and  a 
plane  MN  is  the  point  of  intersection  of  the  line  PE 
and  the  plane.  Thus  if  P  be  any  point,  the  per- 
spective of  P  with  reference  to  the  point  E  and  the 
plane  MN  is  the  point  P'  where  EP  cuts  the  plane 
MN.  Similarly  the  perspective  of  Q  with  reference 
to  the  point  E  and  plane  MN  is  the  point  Q'  where 
EQ  cuts  MN. 

The  point  E  is  the  eye  of  the  observer.  The  per- 
spective of  any  point  P  with  reference  to  any  plane 
MN  is  the  intersection  of  the  line  of  sight  PE  with 
the  plane.  The  plane  MN  is  called  the  perspective 
plane. 

3.  Perspective  of  a  Line. — The  perspective  of 
a  line  will  be  found  by  joining  all  points  of  the  line 
to  the  eye  and  rinding  their  intersections  with  the 
perspective   plane.    The  perspective  of  a   straight 
line  will  be  a  straight  line.    Since  by  joining  P  and 
Q  with  E  we  have  a  triangle  EPQ   (Fig.  i),  and 
the  plane  of  the  triangle  will  cut  the  perspective 
plane  MN  in  a  straight  line  P'Q',  it  is  evident  that  a 
line  joining  any  point  in  PQ  with  E  will  cut  the 
plane  MN  somewhere  in  P'Q',  as  such  a  line  lies  in 
the  plane  EPQ,  and  as  the  plane  cuts  the  perspective 
plane  MN  in  P'Q'.  It  is  sufficient  in  determining  the 
perspective  of  a  definite    part    PQ    of    a    straight 
line,  to  find  the  perspective  of  the  two  points  P  and 
Q  and  join  these  points  by  a  straight  line. 

4.  Point  of  View.  —  A  perspective  is  defined 
with  reference  to  the  point  of  view,  which  is  the  eye 
of  the  observer.     If  we  stand  in  a  room  and  look 
through  a  window  glass  at  a  point  A,  the  intersec- 
tion of  the  line  of  sight  with  the  plane  of  the  win- 


BACKBONE   OF  PERSPECTIVE.  3 

dow  glass  will  be  the  perspective  of  the  point  with 
reference  to  that  particular  location  of  the  eye. 
The  point  A  will  have  a  perspective  for  every  posi- 
tion of  the  eye  in  the  room. 

5.  Perspective  of  a  Point. — Let  P  (Fig.  2)  be 
any  point  and  E  the  eye  and  MN  the  perspective 
plane  which  is  taken  as  vertical.  Let  the  plane 
MN  cut  the  horizontal  plane  in  GL.  Project  P 
and  E  on  the  horizontal  plane  in  p  and  e.  Join  pe, 
cutting  GL  in  D.  The  plane  PpeE  will  cut  plane 
MN  in  a  vertical  line,  as  the  plane  PpeE  is  itself 
vertical.  The  line  of  intersection  of  the  two  ver- 
tical planes  PpeE  and  MN  will  be  parallel  to  Pp 
and  Ee.  The  perspective  of  P  will  lie  somewhere 
on  the  vertical  line  Dt,  and  as  it  must  lie  on  EP  it 
will  be  the  point  where  EP  intersects  this  vertical. 


Figure  2. 


It  is  our  object  to  develop  some  method  of  finding 
this  point  of  intersection. 

Draw  the  perpendiculars  from  P  and  E  to  MN, 


4  BACKBONE   OF  PERSPECTIVE. 

and  let  the  feet  of  these  perpendiculars  be  p'  and  e' 
respectively.  The  projection  of  PE  on  MN  will  be 
p'e'.  The  perspective  t  of  the  point  P  will  be  on 
EP  and  it  will  lie  on  e'p',  because  PE  will  intersect 
the  plane  MN  at  some  point  on  p'e'.  Hence  the  per- 
spective of  the  point  P  will  lie  on  p'e',  and  on  Dt, 
and  therefore  at  their  intersection. 

6.  Perspective  Plane  in  H. — If  in  Fig.  3  we 
revolve  the  plane  MN  about  GL  as  an  axis,  each 
point  in  the  plane  will  describe  a  circle  whose  cen- 
ter lies  on  and  whose  plane  is  perpendicular  to  GL. 
If  the  plane  is  folded  to  the  horizontal  position,  the 
points  p'  and  e'  will  fall  as  far  from  GL  as  the 
points  P  and  E  are  above  H.  Thus  p'  will  de- 
scribe a  circle  whose  radius  is  Qp'  and  whose  center 


Figure  3. 


is  Q.  When  MN  is  folded  into  the  horizontal  posi- 
tion p'  will  fall  at  P!.  Similarly  e'  will  fall  at  Ex 
and  t  will  fall  at  Tx. 

7.     Example.— In   Fig.  3  let  SQ=4",  Qp=z", 


BACKBONE   OF  PERSPECTIVE. 


Figure  4. 


Find  the  perspective  of 
the  point  P.  Draw  as  in 
f?  Fig.  4  a  line  GL  and 
make  SQ  =  4",  and 
make  Se  =  3",  pQ \  = 
2".  Join  pe,  cutting 
GL  at  D.  When  MN 
is  folded  into  the  hori- 
zontal position  p'  and  e' 
(Fig.  3)  will  fall  at  P, 
and  at  El5  respectively.  The  folded  position  of  the 
perspective  will  lie  on  P^  and  the  folded  position 
of  the  vertical  line  Dt  of  Figs.  2  and  3  will  lie  on 
Dt'. 

PROBLEM  i.     If  SQ=2^",  Qp=2,  QP±— o,  eS= 
2^2",  and  SE±=i",  find  the  perspective  of  P. 

PROBLEM  2.    If  SQ=2",  Qp=2",  QP1=i",  eS= 
3",  and  SEj— 2",  find  the  perspective  of  P. 

8.  General  Method. —  The    method     just   out- 
lined is  perfectly  general,  and  can  be  used  to  find 
the  perspective  of  any  structure,  however  compli- 
cated.   The  plane  of  the  paper  upon  which  the  con- 
struction is  made  represents  H,  and  the  perspective 
plane  (which  in  all  practical  cases  is  vertical)   is 
folded  into  H.    The  horizontal  projection  p  of  the 
point  P  is  located  on  the  paper,  and  the  perspective 
is  found  to  be  at  t,  which  (after  folding)  is  in  the 
same  plane  as  p.  The  points  p'  and  e'  are  the  pro- 
jections of  P  and  E  on  the  perspective  plane,  and 
P!  and  Ej  are  the  folded  positions  of  these  points. 

9.  Translation  of  the  Perspective  Plane. — The 
horizontal  projection  of  points  and  the  folded  posi- 


BACKBONE   OF  PERSPECTIVE. 


tion  of  their  projections  on  the  perspective  plane 
are  generally  above  GL,  as  the  perspective  plane  is 
taken  between  the  eye  of  the  observer  and  the  ob- 
ject. Where  there  are  many  points  the  figure  may 
become  complicated.  To  prevent  confusion  and  to 
keep  the  plan  and  perspective  of  the  points  separate, 
the  perspective  plane  before  folding  is  brought  from 
position  GL  in  Fig.  5  to  position  G'L',  while  the 
point  e  and  the  plan  p  of  the  point  remain 
fixed.  It  is  clear  that  Ej  and  Px  will  occupy  exactly 
the  same  positions  with  respect  to  G'L'  that  they 
did  with  respect  to  GL;  that  is,  S'E"=SE1,  Q'P"= 
QPr  The  perspective  T±  of  the  point  P  lies  on  the 
perpendicular  at  D  to  GL  and  on  P^;  and  it  is 
clear  that  in  bring- 
ing GL  to  G'L'  we 
have  simply 
brought  all  the 
points  except  the 
horizontal  projec- 
tion p  into  the  po- 
sition P'^E",  by  aG' 
motion  of  transla- 
tion which  has  not 
altered  their  relative  positions. 

10.  True  Height  Line  of  a  Point.— From  Fig. 
5  we  see  that  the  line  pPx  is  perpendicular  to  GL, 
and  that  Q'P"  is  equal  to  the  true  height  of  the 
point  above  H,  and  that  the  point  P"  is  joined  to 
E".  This  can  be  expressed  in  the  following  con- 
structive rule: 

( i )    Join  the  plan  of  the  point  to  the  plan  of  the 


Figure  5. 


BACKBONE   OF  PERSPECTIVE.  7 

eye,  cutting  GL  in  D,  and  at  D  draw  DD'  perpen- 
dicular  to  the  ground  line. 

(2)  Drop  a  perpendicular  from  p,  the  plan  of  the 
point,  on  the  new  ground  line  (G'Lr)  and  from  G'L' 
lay  off  on  this  perpendicular  the  true  height  of  the 
point=Q'P".  Join  the  point  thus  located  to  E" ,  and 
where  the  line  E"P"  cuts  DD'  is  the  perspective  of 
the  point  P. 

In  the  following  articles  the  perspective  of  the 
point  P  will  be  marked  P',  that  of  K  will  be  marked 
K',  etc. 

ii.  Perspective  Triangle. — Two  points  P  and 
K  whose  heights  above  H  are  O'P"  and  Q'K"  re- 
spectively lie  on  a  common  vertical  line.  Let  the 
height  of  E— S'E"  and  e  and  p  be  located  with  re- 
spect to  GL  and  G'L'  as  in  Fig.  6.  Join  pe,  cutting 


G3 


GL  at  D  and  on  the  true  height  line  Qp  of  P  and  K 
lay  off  Q'K"  equal  to  the  height  oF  K  and  Q'P" 
equal  to  the  height  of  P.  Take  S'E"  equal  to 
height  of  E  and  join  E"P"  and  E"K",  cutting  DD' 


8  BACKBONE   OF  PERSPECTIVE. 

in  K'  and  F.  The  triangle  E"FK'  is  called  the 
perspective  triangle,  and  it  is  the  perspective  of  the 
triangle  EPK. 

PROBLEM  3.  Given  SQ=3",  Q'P"=i",  Q'K"= 
3",  S'E"=2",  find  the  perspective  of  PK. 

12.  Special  Case. — When  the  plan  of  a  point 
lies  on  or  near  eE",  the  foregoing  method  of  article 
10  is  indeterminate  or  lacks  exactness.  In  the  first 
case  the  former  method  can  not,  and  in  the  latter 
case  should  not,  be  used.  In  the  first  case  the 
perspective  triangle  resolves  into  the  straight 
line  eE",  and  in  the  latter  case  its  sides  cut  DD'  at 
such  sharp  angles  that  the  solution  is  not  definite. 


Figure  7. 


In  Fig.  7  let  p  and  e  be  the  plans  of  the  point  P 
and  of  the  eye  E.  If  the  trapezoid  PEep  be  folded 
around  pe  into  H,  PE  will  fall  at  P2E.,,  and  E2e, 
Dtj  and  P2p  will  be  perpendicular  to  pe,  and  Dtt  is 


BACKBONE   OF  PERSPECTIVE.  9 

equal  to  the  height  of  the  perspective  of  P  above 
GL. 

In  Fig.  8  draw  lines  at  e,  D, 
and  p  perpendicular  to  pe  and 
make  eE2  equal  to  the  height 
of  the  eye,  and  P2p  equal  to  the 
height  of  the  point.  Join  P2E2, 
cutting  Dtj  in  t±.  Lay  off  D'P'  p'  s\ 
equal  to  Dt^  The  point  P'  is 
the  perspective  of  P. 

PROBLEM  4.  If  Fig.  5  SQ=o, 
Qp=2",  Q'P"=i"  eS=2",  and  Figure 

S'E2=3",  find  the  perspective  of  P. 

PROBLEM  5.    If  SQ=o,  Qp=i",  Q'P"=o, 
2",  and  S'E2=i",  find  the  perspective  of  P  (Fig.  5). 

PROBLEM  6.  The  plan  of  a  point  whose  height 
above  H  is  3"  is  2"  behind  GL,  and  the  plan  of  the 
eye  whose  height  is  I  inch  is  5  inches  from  GL.  If 
pe  is  at  right  angles  to  GL,  find  the  perspective  of 
P. 

13.  Perspective  of  a  Block. — Let  ABCD  (Fig. 
9)  be  the  base  of  a  rectangular  block,  whose  side 
and  end  views  are  shown  by  ABFT  and  ADHT. 
The  block  rests  on  the  horizontal  plane  and  has  its 
edge  in  the  perspective  plane.  We  have  AB= 
15',  ADr=9',  height  of  blocfc=io',  baL=3O°.  The 
eye  is  taken  25'. in  front  of  the  perspective  plane  and 
at  a  height  of  6'. 

To  find  the  perspective  of  AT  join  ea,  cutting  GL 
at  a  and  from  this  point  drop  a  perpendicular  to 
G'L',  cutting  it  at  A'.  As  AT  lies  in  the  perspec- 
tive plane,  it  will  be  its  own  perspective,  and  all  we 
have  to  do  is  to  lay  off  A'T'=io',  thus  determin- 


10 


BACKBONE   OF  PERSPECTIVE. 


ing  the  perspective  of  AT.  Then  drop  a  perpendic- 
ular from  b  on  G'L',  cutting  it  at  Q'.  Now  as  Q'b 
is  the  "true  height"  line  for  all  points  that  lie  on 

H T 


Figure  9. 

Q'b  lay  off  the  height  of  B  and  F  on  Q'b.  The 
height  of  B  is  zero  and  that  of  F=io'=Q'R'.  Join 
eb,  cutting  GL  at  N.  The  points  B'  and  F'  where 
E"Q'  and  E"R'  cut  the  perpendicular  to  GL  at  N 
are  the  perspectives  of  B  and  F,  respectively.  The 
perspectives  of  DH  and  CK  are  found  as  that  of 
BF.  The  whole  perspective  is  easily  completed,  and 
appears  as  A'B'F'T'H'D'C'K'. 

PROBLEM  7.  Given  (in  Fig.  9)  AB=i2',  AD= 
8',  ea=i6',  height  of  block==8',  height  of  eye=4'. 
The  perspective  plane  GL  passes  i'  below  a  and 
AB  makes  30°  with  GL.  Construct  the  perspective 
of  the  block. 

PROBLEM  8.  Find  the  perspective  of  the  block  of 
Fig.  9  when  AB  makes  45°  with  GL. 


BACKBONE   OF  PERSPECTIVE. 


II 


14.   True  Height  Line  of  a  Horizontal  Line. — 

If  the  line  HK  (Fig.  9)  be  extended  to  the  per- 
spective plane  it  will  intersect  this  plane  directly 
above  G  (or  G')  at  a  distance  equal  to  the  height  of 
HK  above  the  horizontal  plane.  The  true  height 
of  HK  above  the  horizontal  plane  can  be  laid  off  on 
GG'  equal  to  G'J'.  The  point  J'  is  the  perspective 
of  the  point  of  the  line  HK  that  lies  in  the  per- 
spective plane.  Now,  if  we  join  J'H'  we  have  the 
perspective  of  HK  extended  to  the  perspective 
plane.  The  perspective  of  K  lies  somewhere  on  this 
line  and  can  be  found  in  two  ways : 

First.    Join  ek,  and  from  the  point  where  it  cuts 


Figure  9. 


GL  drop  a  perpendicular,  cutting  J'H'  at  K'.  This 
determines^  the  perspective  of  K.  In  the  same  way 
we  can  find  the  perspective  of  any  point  not  in  eE". 


12 


BACKBONE   OF  PERSPECTIVE. 


Second.  Produce  kf  to  cut  GL  at  L.  Then  LL' 
is  the  "true  height"  line  of  all  horizontal  lines  that 
lie  in  the  face  of  BFKC.  Lay  off  L'S'  equal  to  the 
height  (10')  of  line  KF.  Join  S'F'  and  produce  it 
to  cut  J'H'  at  K'. 

One  of  these  methods  should  always  be  used  for 
points  near  eE",  for  the  construction  lines  of  the 
usual  method  intersect  at  such  sharp  angles  that 
their  point  of  intersection  is  not  sufficiently  definite. 
15.  Horizontal  Squares. — A  square  abed,  6'x6', 
lies  on  H  and  its  side  dc,  Fig.  10,  is  parallel  to  and 
i'  from  GL.  The  eye  (whose  height  is  6')  is  10'  in 
front  of  the  perspective  plane  and  lies  in  a  per- 
pendicular to  GL  2'  to  right  of  c.  Find  the  per- 
spective of  abed  and  of  the  nine  squares  into  which 
it  is  divided. 

SOLUTION. — The  true 
height  line  of  ad  is  H'd, 
and  as  D  and  A  lie  on 
the  horizontal  plane, 
their  true  heights  are 
zero.  We  therefore  join 
ea  and  ed  cutting  GL  in 
K  and  P,  respectively. 
The  perspective  of  D 
will  lie  on  E"H'  verti- 
cally below  P,  and  that 
of  a  will  lie  on  E"H' 
£/  vertically  below  K.  The 
perspectives  of  a  and  d 
t  A' 


H 


Figure  10. 


D'.  In  the  same  way  we  find  the  perspectives  of  b  and 
c  at  B'  and  C'.      Join    A'B',  B'C,  CD'  and  D'A', 


BACKBONE   OF  PERSPECTIVE.  13 

forming  A'B'C'D',  the  perspective  of  abed.  The 
perspective  of  any  point  f  can  be  found  by  the  usual 
method. 

1 6.  Perpendiculars  and  Parallels. — It  will  be 
observed  from  the  preceding  example  that  the  per- 
spective of  all  lines  perpendicular  to  the  perspective 
plane  pass  through  a  common  point  E".  When  the 
perspective  of  a  series  of  parallels  passes  through  a 
common  point  as  E",  they  are  said  to  vanish  at  E", 
and  E"  is  called  the  vanishing  point  of  this  system 
of  parallel  lines.  It  will  be  shown  later  that  all  sys- 
tems of  parallel  lines  have  a  vanishing  point. 

The  point  E"  (the  vanishing  point  of  perpendicu- 
lars) is  called  the  center  of  the  picture.  The  per- 
spectives of  all  horizontal  lines  parallel  to  the  per- 
spective plane  are  parallel  to  GL. 

PROBLEM  9.  Find  the  perspective  of  the  square 
in  Fig.  10  when  eE"  lies  on  BC  produced,  other  data 
remaining  the  same. 

PROBLEM  10.  Find  the  perspective  of  the  same 
when  eE''  bisects  AB. 

PROBLEM  n.  Find  the  perspective  of  the  squares 
when  the  eye  is  2'  to  the  left  of  AD. 

PROBLEM  12.  Find  the  perspective  of  a  regular 
hexagon  of  2'  side  that  lies  on  H  when  one  of  its 
sides  is  parallel  to  and  its  center  is  3'  from  GL, 
when  the  eye  (height  4')  lies  6'  in  front  of  the  per- 
spective plan  and  in  a  perpendicular  from  the  center 
of  hexagon. 

PROBLEM  13.  Find  the  perspective  of  hexagon 
in  problem  12  when  one  side  is  perpendicular  to 
GL. 


I4  BACKBONE   OF  PERSPECTIVE. 

PROBLEM  14.  A  circle  of  diameter  4'  lies  on  H 
and  has  its  center  3'  from  the  perspective  plane. 
Find  the  perspective  of  the  circle  when  the  eye 
(height  5')  lies  in  a  perpendicular  through  center  to 
GL  and  5'  in  front  of  GL. 

PROBLEM  15.  Find  the  perspective  of  a  circle 
whose  plane  is  parallel  to  the  perspective  plane  and 
at  a  distance  of  3'  from  it,  the  height  of  its  center 
above  H  being  5',  when  the  eye  (height  5')  lies  in 


Figure  11. 


the  perpendicular  to  the  perspective  plane  from  the 
center  of  the  circle  and  6'  in  front  of  the  perspec- 
tive plane.  Diameter  of  circle=4'. 


BACKBONE   OF  PERSPECTIVE.  15 

17.  Perspective  of  Steps.  —  A  series  of  four 
steps  (Fig.  ii )  six  feet  long,  each  having  a  tread 
and  rise  of  i',  has  one  end  parallel  to  the  perspec- 
tive plane  at  a  distance  i'  from  it.  Find  the  per- 
spective of  the  steps  when  the  eye  (height  8')  is  6' 
to  the  right  of  lowest  step  and  9'  in  front  of  the 
perspective  plane.  Height  of  base  fbc=i'. 

On  aA"  the  true  height  line  of  ad  lay  off  the 
heights  of  the  different  steps  from  G'L'  equal  to  I, 
2,  3,  and  4  feet  at  N',  R',  S',  T  and  Q'.  Join  ea 
and  ed,  cutting  GL  at  P  and  I.  Drop  perpendicu- 
lars from  P  and  I,  cutting  E"Q'  in  A'  and  D', 
the  perspectives  of  a  and  d.  The  perspective  of  f 
lies  on  E"N'  and  vertically  below  P.  Join  R',  S', 
and  T'  to  E",  cutting  A'F'  at  M',  etc.,  through 
which  draw  lines  parallel  to  G'L'.  To  find  the  per- 
spective B'X'  of  an  edge  BX,  join  eb,  and  from  the 
point  where  it  cuts  through  GL  drop  a  perpendicular 
to  cut  the  horizontals  through  F'  and  M'  at  B'  and 
X'.  Join  B'  and  X'  to  E"  and  drop  a  perpendicular 
from  point  of  intersection  of  ec  and  GL,  cutting 
E"B'  at  C',  the  perspective  of  c.  In  the  same  way 
we  can  find  the  perspective  of  all  other  edges. 

PROBLEMS. 

PROBLEM  16.  Find  the  perspective  of  the  steps 
in  Fig.  ii  when  height  of  eye  is  9'  and  lies  4'  to 
the  right  of  the  lowest  step,  other  dimensions  re- 
maining the  same. 

17.  Find  the  perspective  of  a  hollow  box  4'x6', 
height  5',  that  rests  on  H  with  the  long  face  per- 
pendicular to  the  perspective  plane.  The  open  end 
next  the  perspective  plane  is  2'  from  it,  and  the  eye 


16  BACKBONE   OF  PERSPECTIVE. 

(height  3')  lies  on  a  line  bisecting  the  plan  of  the 
box  and  at  a  distance  of  10'  from  the  perspective 
plane. 

18.  Find  the  perspective  in  problem  17  when  the 
height  of  the  eye  is  i',  other  dimensions  remaining 
the  same. 

19.  One  corner  of  the  box  in  problem  17  lies 
in  the  perspective  plane  and  the  short  end  makes  an 
angle  of  30°  with  it.    If  the  eye  (height  2^')  lies  in 
a  perpendicular  to  GL  at  the  nearest  corner,  find  the 
perspective  of  the  box. 

20.  The  base  of  a  monument  is  4'x6'  by  i'  high. 
Upon  the  base  rests  a  rectangular  block  3^5'  by  3' 
high.     The  block  is   capped  by  a  pyramid  whose 
base  is  the  upper  base  of  the  block  and  whose  height 
is  2'.    The  longest  face  of  the  base  makes  30°  with 
the  perspective  plane   and  the   nearest  corner   of 
base  is  i'  from  the     perspective     plane.     The  eye 
(height  2')  lies  in  a  perpendicular  from  the  nearest 
corner  of  base  and  is  8'  in  front  of  the  perspective 
plane.    Find  the  perspective. 

21.  Find  the  perspective  in  problem  20  when  the 
height  of  the  eye  is  5',  other  dimensions  remaining 
fixed. 

22.  Find  the  perspective  in  problem  20  when  the 
long  face  of  base  makes  45°   with  the  perspective 
plane,  other  dimensions  remaining  as  in  problem  20. 

23.  Find  the  perspective  in  problem  20  if  the 
height  of  the  eye  is  i',  other  dimensions  remaining 
the  same. 

24^  Eight  cubes  are  the  corners  of  a  larger  cube 
of  12'  edge.  One  face  of  the  cubes  is  parallel  to  the 
perspective  plane  and  2'  from  it.  The  eye  (height 


BACKBONE   OF  PERSPECTIVE.  17 

6')  lies  in  a  perpendicular  to  GL  from  the  central 
line  of  the  plan  and  is  12'  from  the  perspective 
plane.  Draw  the  perspective  of  the  cubes  if  edge 
=2'. 

25.  A  framework  whose  outer  dimensions  form 
a  cube  12'  edge  is  composed  of  pieces  i'xi'  along 
each  edge  of  the  cube.    One  corner  of  the  cube  lies 
in  the  perspective  plane  and  the  plane  of  one  face 
makes  30°  with  it.     If  the  eye  (height  6')  lies  in  a 
perpendicular  to  GL  through  the  corner  in  the  per- 
spective plane  and  is  15'  from  it,  draw  the  perspec- 
tive of  the  framework. 

26.  Draw  the  perspective  of  the  framework  in 
problem  25  when  the  eye  is  moved  2'  to  the  right, 
other  dimensions  remaining  the  same. 

27.  Draw  the  perspective  of  the  framework  in 
problem  25  when  the  height  of  the  eye  is  4',  other 
dimensions  remaining  the  same. 

28.  Draw  the  perspective  of  the  framework  in 
problem  25  when  the  front  face  of  the  cube  lies  in 
the  perspective  plane  and  the  eye  (height  8')  lies  in 
the  central  line  of  plan  that  is  perpendicular  to  GL. 
Other  dimensions  are  the  same  as  in  problem  25. 


CHAPTER  TWO. 

Vanishing  Point  Method. 

1 8.  Vanishing  Points. — To  find  the  perspective 
of  a  line  AB  in  Fig.  12  whose  height  above  H  is  h, 
we  can  find  the  perspective  of  A  and  B  as  before. 
But  if  one  of  the  points  is  on  eE"  or  near  it  the  solu- 
tion by  the  former  method  lacks  exactness.  It  is 


Figure  12. 

advisable  to  find  the  perspective  of  points  on  AB 
favorably  located.  One  of  these  favorable  points 
is  K  where  AB  cuts  the  perspective  plane  at  a  height 
h  above  GL.  The  second  favorable  point  should  be 
taken  as  far  from  eE"  as  the  limits  of  the  paper  will 
permit.  Let  Q  be  such  a  point.  The  perspective 

18 


BACKBONE   OF  PERSPECTIVE.  ig 

Q'  of  Q  is  found  as  follows  :  Drop  a  perpendicular 
from  q  on  G'L',  cutting  it  at  P'  ;  lay  off  P'Q"  equal 
to  the  true  height  of  point  Q.  Join  Q"  and  E"  ;  join 
eq  cutting  GL  at  D  and  draw  DD'  perpendicular  to 
GL,  cutting  E"Q"  at  Q',  the  perspective  of  Q.  K'Q' 
is  the  perspective  of  the  line  desired.  The  perspec- 
tives of  A  and  B  are  found  by  joining  A  and  B  to 
e  and  by  dropping  perpendiculars  from  the  points 
where  these  lines  cut  GL  to  cut  the  line  K'Q'. 

The  line  E"F  is  drawn  parallel  to  GL  and  G'L'. 

In  the  similar  triangles  E"SQ'  and  E"FQ", 


FQ"     E"F* 
But  E"S=OD',  E"F=OP'. 


= 

FQ"     OP7' 

Let  t  =  height  of  eye  above  H  =  OE". 
h  =  height  of  AB  above  H  =  P'Q". 
x  =  OP'  =  distance  from  eye  to  the  true 

height  line  of  Q. 

z  =  OD'  =  distance  from  eye  line  to  the  ver- 
tical through  the  perspective  of  the 
point. 

y  =  SQ'  =  height  of  perspective  of  point 
from  horizontal  through  E",  called  the 
horizon. 


Now,  let  Q  move  along  AB  to  an  infinite  distance 
from  K.     The  line  eq  that  is  drawn  from  q  to  e 


20  BACKBONE   OF  PERSPECTIVE. 

will  then  be  parallel  to  AB  and  will  cut  GL  at  L, 
and  D'  will  move  to  L'.    But  x  becomes  infinity, 


Therefore  the  perspective  of  the  point  on  AB  that 
is  at  infinity  is  zero  distance  from  E"V  in  LL'.  The 
line  E"V  is  called  the  horizon. 

Corollary  :  The  perspectives  of  all  lines  parallel  to 
AB  pass  through  V,  and  are  said  to  vanish  at  V. 

From  these  results  we  derive  the  following  rule 
for  obtaining  the  vanishing  point  of  a  system  of 
parallel  lines: 

Rule.  —  Through  the  plan  (e)  of  the  eye  draw  a 
line  parallel  to  the  lines  zvhose  vanishing  point  is 
desired.  From  the  point  where  this  line  cuts  the 
perspective  plane,  GL,  drop  a  perpendicular  to  cut 
the  horizon  at  V. 

Thus  the  vanishing  point  is  absolutely  indepen- 
dent of  the  height  of  the  line  above  H  and  of  the 
position  of  its  plan.  The  only  controlling  factors 
that  locate  the  vanishing  point  are  the  angle  the 
system  of  lines  makes  with  GL  and  the  position  of 
the  eye. 

eg.  Perspective  of  Cross.  —  Given  the  side  and 
end  views  of  a  cross  and  base  as  shown  in  Fig. 
13.  Let  abed  be  the  plan  of  the  base  and  let  the 
perspective  plane  pass  through  the  corner  a. 
Through  e  draw  lines  parallel  to  the  sides  of  the 
base,  cutting  GL  at  G  and  L.  From  these  points 
draw  perpendiculars  to  GL,  cutting  the  horizon 
line  at  V  and  V.  As  the  corner  of  the  base  is  in 
the  perspective  plane,  lay  off  A'F'  equal  to  the 


BACKBONE   OF  PERSPECTIVE. 


21 


height  of  base  and  join  A'  and  F'  to  V  and  V. 
Join  b  and  d  to  e,  cutting  GL  at  i  and  4,  and  from 
these  points  draw  perpendiculars  to  GL,  cutting 
F'V  and  F'V  at  B'  and  D'.  The  base  is  thus 
defined. 

To  draw  the  perspective  of  the  arms  of  the  cross 
produce  one  side  to  cut  the  perspective  plane  at  M. 


Figure  13. 

Then  MM'  is  the  true  height  line  for  all  lines  in 
this  face  of  the  arm.  Lay  off  M'J'  and  M'H'  equal 
to  the  true  heights  of  the  upper  and  lower  surfaces 
of  the  arm.  Join  J'V  and  H'V.  Join  5  and  6  to  e 
and  from  the  points  where  these  lines  cut  GL  draw 
perpendiculars  to  GL.  The  intersection  of  these 
perpendiculars  with  J'V'  and  H'V'will  define  the 
front  face  of  the  arm. 

To  draw  the  upright,  lay  off  the  height  of  the 
base  equal  to  M'Q'  and  height  of  top=M'R'  on  M'M 


22  BACKBONE   OF  PERSPECTIVE. 

from  M'.  Join  these  to  V  and  join  s,  k  and  t  to  e, 
and  from  the  points  of  intersection  of  these  lines 
with  GL  draw  perpendiculars  to  GL,  cutting  R'V 
in  K'  and  S'.  Draw  K'  V  and  join  t  to  e  and  from 
point  where  et  cuts  GL  drop  perpendicular  to  cut 
K'  V  at  T1.  The  rest  can  be  drawn  in  the  same 
way. 

PROBLEM  29.  Construct  the  perspective  of  the 
cross  in  Fig.  13  when  side  of  base  =  7',  end  of 
base  =  5',  length  of  crossarm  =  6',  other  dimen- 
sions as  in  Fig.  20. 

20.  Perspective  of  a  House. — Let  one  corner  a 
of  the  plan  abed  (Fig.  14)  be  in  the  perspective 
plane.  Through  e  draw  lines  parallel  to  ab  and  ad, 
cutting  GL  in  L  and  G.  From  these  points  drop 
perpendiculars,  cutting  the  horizon  in  V  and  V, 
the  vanishing  points  for  the  systems  of  lines  parallel 
to  ab  and  ad,  respectively.  Draw  a  line  from  a  to  e. 
The  point  A'  where  it  cuts  G'L'  will  be  the  perspec- 
tive of  a.  Join  A'V  and  A'V  and  from  the  points 
where  eb  and  ed  cut  GL  drop  perpendiculars  cut- 
ting A'V  and  A'V  in  B'  and  D',  the  perspective  of 
the  points  b  and  d.  Lay  off  the  true  heights  of  all 
points  in  the  visible  sides  of  the  house  on  vertical 
A'a  from  G'L',  and  the  perspective  of  these  points 
will  lie  on  the  line  joining  their  proper  height  to 
their  vanishing  point.  To  draw  the  window  pq,  lay 
off  the  height  of  the  top  and  bottom  of  window 
from  A'  on  aA'  at  K'  and  T'.  Join  T'  and  K'  to  V, 
and  from  the  points  where  ep  and  eq  cut  GL  drop 
perpendiculars  to  cut  K'V  and  TV  in  P',  M',  Q', 
and  O'.  To  locate  the  roof,  extend  the  comb  to  cut 
the  GL  at  U,  and  drop  a  perpendicular  from  U  on 


BACKBONE   OF  PERSPECTIVE.  23 

G'L',  cutting  it  at  U'.  Lay  off  on  UlT  from  U'  the 
true  height  of  the  comb,  equal  to  U'x,  and  join  x  to 
V.  Join  the  ends  of  the  comb  to  e,  and  where  these 
lines  cut  GL,  drop  perpendiculars,  cutting  the  line 
xV  in  Z'  and  Y',  the  perspective  of  the  ends  of  the 
ridge.  To  find  the  perspective  of  the  eaves,  extend 
the  plan  of  the  eave  line  to  cut  GL  say  at  N,  and 


Figure  14. 

from  N'  lay  off  true  height  of  eave  line  equal  to 
N'F',  and  join  F'  to  V.  From  the  points  where  er 
and  es  cut  GL,  drop  perpendiculars,  cuting  F'V  in 
R'  and  S'  the  perspectives  of  r  and  s.  In  the  same 
way  the  perspectives  of  the  other  eave-lines  can  be 
found. 

PROBLEM  30.  Construct  the  complete  perspective 
of  the  house  in  Fig.  14  when  AB  =  6',  AD  =  4', 
height  of  eaves  =  4',  height  of  comb  =  6j^',  height 
of  eye  =  2',  distance  from  eye  to  perspective  plane 
=  12',  other  dimensions  remaining  as  in  Fig.  14. 


24  BACKBONE   OF  PERSPECTIVE. 


BACKBONE   OF  PERSPECTIVE.  25 

21.  Architectural  Perspective. — In  finding  the 
perspective  of  a  house,  it  is  convenient  to  let  one 
corner  of  the  house  pass  through  or  lie  in  the  per- 
spective plane.  Thus,  if  is  it  desired  to  construct 
the  perspective  of  the  house  whose  side  and  front 
elevations  are  marked  "front"  and  "side"  (Fig.  15), 
it  will  shorten  the  work  if  we  let  a  corner  common 
to  the  two  views  given  lie  in  the  perspective  plane 
GL.  It  is  always  best  to  take  e  as  low  as  the  size 
of  the  drawing  sheet  will  permit.  After  e  is  locat- 
ed draw  lines  through  e  parallel  to  the  sides  of  the 
house,  cutting  GL  in  A  and  B,  and  then  drop  per- 
pendiculars from  A  and  B  to  the  horizon  line 
through  E",  cutting  the  horizon  in  V  and  V.  The 
corner  that  lies  in  the  perspective  .plane  will  be  the 
true  height  line  for  all  points  lying  in  the  planes  of 
the  two  faces  or  sides  seen  in  the  two  views.  These 
heights  are  laid  off  from  G'L'  on  verticals  through 
E"  and  joined  to  V  or  V.  To  find  the  perspective 
of  any  horizontal  line  like  the  ridge  or  comb  pq, 
we  produce  pq  to  cut  the  perspective  plane  at  F 
and  from  F  drop  a  perpendicular  cutting  G'L'  at  F'. 
On  line  F'F  lay  off  the  true  height  of  the  ridge  equal 
to  F'D.  Join  D  and  V,  and  join  e  and  p,  cutting 
GL  at  G,  and  from  G  drop  a  perpendicular,  cutting 
DV  at  P',  which  is  the  perspective  of  p,  or  the  left 
end  of  ridge.  In  the  same  way  we  can  find  T',  the 
perspective  of  t.  Join  T'  to  V  and  where  it  cuts 
DV  will  be  the  perspective  of  q.  Other  horizontal 
lines  can  be  found  in  the  same  way.  If  a  line  is 
not  horizontal  in  the  building  it  is  best  to  find  the 
perspective  of  each  end  separately  by  the  projective 
method. 


CHAPTER  THREE. 


Axometric  Projections. 

22.  In  order  to  show  the  different  parts,  con- 
nections and  relations  of  a  framework,  it  is  often 
desirable  to  take  its  projection  on  some  plane  not 
parallel  to  any  of  the  plane  faces.    The  basal  planes 
of  most  frameworks  are  composed  of  a  series  of 
surfaces  each  of  which  is  at  right  angles  to  the 
other  two.     Three  axes  each  at  right  angles  to  the 
other  and  parallel  to  the  edges  of  the  framework 
can  be  drawn,  and  all  lines  can  then  be  located  with 
references  to  these  axes. 

23.  Axes.— Let  PA,  PB,  and  PC  in  Fig.  16  be 
the  axes  in  space  each  at  right  angles  to  the  other 
two,  and  let  these  axes  be  _ 
projected      normally      on 

any  plane  ABC  in  KA, 
KB,  and  KC.  Let  PAK 
=  a,  PBK  =  b,  PCK  = 
c.  Let  AK,  BK,  and  CK 
be  produced  to  intersect 
BC,  AC,  and  AB  in  D,  E, 
and  F.  The  lines  KA, 
KB  and  KC  are  the  axes 
of  projection. 

Then  as  PA  and  PK  are  at  right  angles  respec- 
tively to  the  planes  BPC  and  ABC,  the  plane  of 
these  two  lines  is  at  right  angles  to  the  intersection 
of  the  two  planes  BPC  and  ABC.  That  is,  APD 
is  at  right  angles  to  BC.  Hence  AD  and  PD  are 

26 


BACKBONE   OF  PERSPECTIVE,  27 

the  altitudes  of  the  triangles  ABC  and  BPC.    Sim- 
ilar relations  are  true  of  BE  and  CF. 

The  right  angles  BPC,  APC,  and  APB  are  pro- 
jected in  the  angles  BKC,  AKC,  and  AKB,  which 
we  represent  by  x,  y,  and  z. 

24.  Reduction  Cosines.  —  The  axis  PA  has  been 
projected  in  KA  and  its  length  has  been  reduced 
from  PA  to  KA.     But  in  the  right  triangle  PAK 
we  have 

KA  =  PA  cos  a. 
Similarly 

KB  =  PB  cos  b. 
KC  =  PC  cos  c. 

Thus  each  axis  is  reduced"  in  the  ratio  of  the 
cosine  of  the  angle  it  makes  with  the  plane  of  pro- 
jection. All  lines  that  are  parallel  to  PA,  PB,  and 
PC  will,  when  projected  on  the  plane  ABC  be 
reduced  in  the  ratio  of  the  cosines  of  a,  b,  and  c,  and 
these  are  therefore  called  the  reduction  cosines. 

25.  Fundamental  Formula.  —  In  the  right  tri- 
angles PBK  and  PCK  of  Fig.  16  we  have 

PK       h 


PK  __  h_ 
~  PC  ~PC 

where  PK  =  h 

Squaring  and  adding, 

h2        h2          PB2+PC2 

'  ' 


In  the  right  triangle  PBC  we  have 
PB2       PC2  =  BC2 


28  BACKBONE   OF  PERSPECTIVE. 

Also  as  expressions  for  the  area  M  of  PBC  we 
have  the  following  : 

PBxPC     PDxBC 


or  PB2PC2  =  PD2BC2 
Substituting  in  (A)  gives 

BC2  h2 

=  h2pD^B^  =  PD*  .....  (B) 

But  in  the  right  triangle  APD,  DPK  =  PAK=a, 
and  cos  DPK  =  —  =  cos  a>    Putting  this  value  of 

(B)  we  get 


sin2b  +  sin2c  =  cos2a 
.  •  .  sin2a  -j-  sin2b  +  sin2c  ==  i 
or,  cos2a  +  cos2b  +  cos2c  =  2  (i) 

26.     Angles   between  Axes.  —  In   the  triangle 
CKD  of  Fig.  16  we  have 


(C) 


Now  DKC  —  180°  —  y,  and  in  the  right  triangles 
PKD  and  PKC, 

DK  =  PK  tan  DPK  =  h  tan  a 

CK          PK  h 


tan  PCK      tan  c 

Substituting  these  values  of  PK  and  CK  in  (C) 
we  get 

cos  ( 180°  —  y )  =  tan  a  tan  c 
.  • .  cos  y  =  —  tan  a  tan  c 


OF 


^BACKBONE 


OF  PERSPECTIVE. 
Similarly 

cos  x  =  —  tan  b  tan  c 
cos  z  =  —  tan  a  tan  b 
PROBLEM  31.    In  the  right  triangle  DKC,  sin2y  = 
DC2      PC2-PD2   .  h       _^         h 

\^t  — .    Jr  LJ    =  

sin  c  cos  a 


(2) 


sin  b 


KC2          KC2 

and  KC  =  h  cot  c,  then  show  that  cos  y 

cos  a  cos  c 

PROBLEM   32.     Given   cos  a  =  2/3,  cos   b  =  y2, 
find  the  cosines  of  x,  y,  and  z. 

27.  Relation  of  Reduction  Angles. — It  is  as- 
sumed that  the  axes  make 
some  definite  angles  with 
the  plane  of  projection. 
These  angles  must  lie  be- 
tween o°  and  90°.  As 
neither  of  the  cosines  can 
be  o  or  unity,  it  follows 
that  the  square  of  either 
cosine  must  be  less  than 
unity;  and  therefore  that 
the  sum  of  the  squares  of  Figure  17. 

the  other  two  cosines  must  be  greater  than  unity. 
If  a  =  32°,  then  90  —  a  =  58°.  Now  if  we  assume 
that  b  =  58°,  we  have 

cos2  a  +  cos2  b  =  cos2  32°  +  cos2  58°  = 

cos2  32°  +  sin2  32°  =  i. 

But  as  cos2a  -f  cos2b  +  cos2c  =2, 

then  cos2c  =  i , .  • .  cos  c  =  I ,  . • .  c  =  a 
This  is  contrary  to  hypothesis.  Now  as  the  cosine 
of  an  angle  decreases  as  the  angle  increases,  we  see 
that  for  possible  values  b  must  be  less  than  58°.    It 


BACKBONE   OF  PERSPECTIVE. 


can  have  any  value  between  o  and  58°.  Thus  none 
of  the  angles  a,  b,  or  c  can  be  equal  to  or  greater 
than  the  complement  of  either  of  the  others. 

Case  I :      If  a  =  90°—  b,  then  cos2a  +  cos2b  =  1 

.'.  cos2c  =  1  .'.  c  =0°. 

One  axis  is  parallel  to 
the  plane  of  projection 
and  the  plane  of  the  oth- 
er two  perpendicular  to 
the  plane  of  projection. 
This  is  the  case  of  ordi- 
nary plans  and  eleva- 
tions, and  is  not  treated 
here.  Figure  IT. 

Case  II:  b>90°-a.  Then  cos'a  +  cos2b<l 
."-.  cos2c>l.  c  is  impossible. 

Case  III:  If  b<90°  —  a,  then  cos2a  +  cos2b>l 
.'.  cos2c<l,  c  is  possible. 

28.  Application. — If  we  assume  the  angles  a, 
b,  and  c,  then  each  dimension  of  the  structure  paral- 
lel to  PA,  PB,  and  PC  will  be  reduced  in  the  ratio 
of  the  cosine  of  a,  b,  and  c. 

EXAMPLE. — Let  cos  a  =  Jv's",  cos  b  =  Vf  ,  then 

cos  c  =  v/|' . 
sin  a  =  J,  sin  b  =  Vf ,  sin  c  =V^  . 

tan  a  =  —^ ,  tan  b=v/r,  tan  c  =»v i  . 

cos  x  =  —tan  b  tan  c  =  —  V^. 
cos  y  =  —tan  a  tan  c  =  —  v^£. 
cos  z  =  —tan  a  tan  b  =  —  v~f . 


BACKBONE   OF  PERSPECTIVE.  31 

log  cos  x  =  l/2  [log  5  —  log  21  ] 

=  l/2    [0.698970  —  I.3222I9] 

=  l/2  [19.376751  —  20]  =  9-6883755 
.-.X=  119°  12' 21" 

log  cos  y  =  y2  [log  i  —  log  21  ] 
=  'l/2  [0—1.322219] 

=    l/2    [18.677781—20]    =  9.3388905 

.•.y=i02°36'i6" 

log  cos  z  =  y2  [log  5  —  log  9] 

=    l/2    [0.698970—  .954243] 
=   V*    [19744727  —  20]    =9.8723635 
.  -.2=138°    II'  23" 

Check  :  x  +  y  +  z  =  360° 

Now,  all  dimensions  of  the  structure  will  be  re- 
duced in  the  ratios  of  cos  a  (=  .8660),  cos  b 
(•=.  .6124),  and  cos  c  (  =  -9354),  and  a  separate 
scale  will  have  to  be  made  for  each  axis. 

29.  Reduction  Ratios. —  Instead  of  assuming 
the  angles,  we  can  assume  their  ratios.  Thus 

Let  cos  a  =  lp,  cos  b  =  mp,  cos  c  =  np. 

cos  a :  cos  b :  cos  c  : :  1 :  m :  n. 

Then  from  (i) 

J2p2   +  m2p2   +  n2p2  =  2 


2l2  ,,  2m2 

cos2a  =  r^ ; ; ,  cos2b  = 


2n2 

COS2C  = 


From  these  we  can  find  tan  a,  tan  b,  and  tan  c,  and 
then  x,  y,  and  z. 


32  BACKBONE   OF   PERSPECTIVE. 

30.  Practical  Application. — We  can,  instead  of 
using  the  incommensurable  fractions  represented  by 
the  cosines  of  a,  b,  and  c,  use  their  ratios,  1,  m,  and 
n,  and  thus  get  the  relative  dimensions.    The  object 
of  axometric  projections  is  to  show  the  connections 
and  relations  more  fully  and  completely  than  ordi- 
nary projections  or  perspectives  would  do.     It  is 
therefore  allowable  to  vary  these  ratios  proportion- 
ally.    It  will  be  far  more  expeditious  and  far  more 
satisfactory  to  assume  the  ratios  1,  m,  and  n,  and  find 
the  values  of  x,  y,  and  z  from  these. 

31.  Systems. —  There  are  three  general   sys- 
tems: 

Isometric :    1  :  m  :  n,  where  1  =  m  —  n ;  all  equal. 
Dimetric  :    1  =  m  or  n,  or  m  =  n ;  two  equal. 
Trimetric :    1  :  m  :  n  ;  all  different. 
In  any  system  the  ratios  must  fulfill  formulas  ( i ) 
and  (2). 

32.  Isometric  System. — The  ratios  are  1  :  m  :  n 
—  i :  i  :  i. 

2 

P      =  12    .—2    .   -2  =  ?    ' 


cos2a  =  2/3,  cos2b  =  2/3,  cos2c  =  ft. 

sin2a  —  ft,  sin2b  =  ft,  sin2c  =  ft. 

tan2a  =  y2,  tan2b  =  ft,  tan2c  =  ft. 

cos  x  =  —  tan  b  tan  c  =  —  ft, .  • .  x  — :  120°. 

cos  y  =  —  tan  a  tan  c  =  —  ft, . ' .  y  =  120°. 

cos  z  =  —  tan  a  tan  b  —  —  ft, . ' .  z  =  120°. 

33.    Isometric  of  Stone  Cap. — In  the  isometric 
system,  the  line  OX  is  drawn  vertically,  and  then 


BACKBONE   OF  PERSPECTIVE. 


33 


the  angles  XOA  and  XOB  are  made  equal  to  120°, 
thus  making  AOB  120°.  Let  OH  equal  the  total 
length  of  a  stone  cap,  OB  the  total  width  and  HG 
the  total  height.  Make  OB  =  its  true  width,  and 
OC  equal  to  its  true  length,  and  draw  the  parallelo- 
gram OBDC.  Then  draw  CF  and  DE  parallel  to 
axis  OA  and  make  CF  equal 
to  its  true  length,  and  draw 
FE  parallel  to  axis  OB.  To 
locate  the  top  GMLK, 
through  G  draw  lines  paral- 
lel to  the  axis  OA  and  OB, 
and  make  GM  and  GK  equal 
to  their  true  lengths  in  the 
stone  cap,  and  draw  lines 
GK,  ML,  and  KL.  Then 
join  KF  and  EL,  forming 
the  parallelogram  FKLE.  To  draw  the  back  face 
AGMR,  make  OA  equal  to  its  true  length  and  join 
AG.  Draw  BR  parallel  to  OA  and  AR  parallel  to 
OB,  intersecting  at  R.  Join  R  with  M,  thus  com- 
pleting the  figure. 

PROBLEM  33.  Draw  an  isometric  of  a  cube  4' 
edge  that  has  a  square  hole  (i'xi')  connecting  each 
pair  of  parallel  faces. 

34.  One-Half  Dimetric  System. — In  the  dimet- 
ric  system  two  of  the  ratios  are  equal,  but  they  can 
bear  a  variety  of  relations  to  the  third. 

The  one-half  dimetric :  (1  :  m  :  n)  —  ( I  \y2  :  I ) 


Figure  18. 


!2+m2 


__  8 


34 


BACKBONE   OF  PERSPECTIVE. 

cos2a  =  |,  cos2b  =  |,  cos2c  =  |  : 
sin^a  =  J,  sin2b  =  },  sin2c  =]. 
tan2a  =  J,  tan2b  =  |,  tan2c  =  £. 
.'.  cos  x  =  —tan  b  tan  c  =  —  J\/7~ 
cos  y  =  —tan  a  tan  c  =  —  J 
cos  z  =  —tan  a  tan  b  =  — 


As  an  example  of  the  application  of  this  system 


Figure  19. 


draw  the  dimetric  of  the  monument  in  problem  20. 

Draw  in  Fig.  19  a  vertical  line  and  make  OA  =  i, 
and  draw  AM  at  right  angles  to  OA.  With  O  as 
a  center  and  8  as  a  radius,  cut  AM  at  M.  Join  OM 
and  produce  in  OC. 

Now  cos  MOA  =   -—  = 


/.  cos  COA  =  -J  /.  COA  =  y. 

As  x  and  z  are  equal,  bisect  the  angle  COA  and 
produce  in  OB,  then  COB  =  x,  and  BOA  =  z.  Full 
dimensions  are  laid  off  on  OA  and  OC  and  one-half 


BACKBONE   OF  PERSPECTIVE.  35 

of  the  real  dimensions  are  laid  off  on  OB.  Thus 
Oa  and  Oc  are  made  equal  to  i'  and  6'  respectively 
—  the  height  and  length  of  the  side  of  the  base  of 
the  monument,  while  Ob  is  laid  off  equal  to  one- 
half  of  the  real  length  (4')  of  the  end  of  the  base. 
Through  a,  b,  and  c  draw  lines  parallel  to  the  axes 
intersecting  in  e  and  f.  Lay  off  ag  and  ke  equal  to 
i'  to  the  full  scale,  while  ah  and  fn  are  laid  off  to 
the  half  scale.  Through  g  and  k  draw  lines  parallel 
to  the  axis  OB,  and  through  h  and  n  lines  parallel 
to  OC,  intersecting  the  former  lines  in  p,  q,  and  r. 
Through  p  draw  a  vertical  (parallel  to  OA)  and 
make  ps  =  3'  (the  height  of  the  block).  Then 
through  s  draw  lines  parallel  to  the  axes  and  make 
su=^5',  and  st  =  3'  on  the  half  scale.  Complete 
the  parallelogram  tsuv,  the  top  of  the  block.  Draw 
the  diagonals  and  from  the  point  of  intersection  3 
lay  off  34  =  2'  and  join  4  to  s,  t,  u,  and  v. 

35.  Three-Fourths  Dimetric.  —  In  this  system 
the  same  scale  is  used  on  two  of  the  axes,  while  a 
scale  equal  to  three-fourths  of  the  first  is  used  on 
the  other  one. 

In  the  three-fourth  system  we  have, 
(1   :  m   :  n)  =  (1  :  J   :  1) 


cos2a  =  ff,  sin2a  =  ¥9T,  tan2a  =  ^; 
Cos2b  =  if,  sin2b  =  -|f  ,  tan2b  =  \  f  ; 
cos2c  =  ff  ,  sin2c  =  T9T,  tan2c  =  -fa. 
cos  y  =  —tan  a  tan  c  =  —  A- 

In  Fig.  20  draw  OA  =  9  parts  on  some  scale, 
draw  AM  perpendicular  to  OA,  and  with  O  as  a 


36  BACKBONE   OF  PERSPECTIVE. 

center  and  a  radius  equal  to  32,  cut  AM  at  M.  Join 
OM  and  produce  in  OC.  Then  bisect  angle  AOC 
and  produce  the  bisector  in  the  line  OB.  Now,  full 
dimensions  are  laid  off  on  the  axes  OA  and  OC  and 
three-fourth  dimensions  on  OB.  On  axis  OC  make 
O2  =  7",  the  full  length  of  the  side  O2.  Now,  the 
side  03  is  equal  in  the  original  figure  to  eight 
inches,  but  we  lay  off  only  three-fourths  of  this  on 


03 ;  that  is,  we  make  03  =  6".  Draw  lines  04,  25, 
and  36  parallel  to  axis  OA  each  equal  to  one  inch, 
and  through  4,  5,  and  6  draw  lines  parallel  to  the 
axes,  marking  out  the  top  of  the  base.  The  shaft 
of  the  cross  is  i"  by  i" '.  Make  distance  48=3", 
and  draw  8g  —  three-fourths  of  3.5".  Then  through 
g  draw  lines  parallel  to  the  axes  and  lay  off  dimen- 
sions equal  to  one  inch  on  axis  parallel  to  OC  and 
to  three-fourths  of  one  inch  on  axis  parallel  to  axis 
OB.  Make  gc  =  4",  be  =  i",  ba  =  i" ,  and  through 
these  points  draw  lines  parallel  to  the  axes.  Make 
ax  =  i",  xy  =  $4",  be  =  2^£",  etc. 


BACKBONE   OF  PERSPECTIVE.  37 

PROBLEM  34.  Draw  the  one-half  dimetric  of  a 
box  without  top,  whose  outside  dimensions  are 
2'x3'x2^',  the  thickness  of  the  material  being  2". 

36.  The  One-Third  Dimetric.—  (1:1/3:1).  We 
have 


•n2  = 


cosa 


2a  =  j-  1  ,  sin2a  =  ^g  ,  tan2a  = 
cos2b  =  T*9,  sin2b  =  }  j,  tan2b  = 
cos2c  =-}!,  sin2c  =  1*9  ,  tan2c  = 
cos  x  =  —tan  b  tan  c  =  — 
cos  y  =  —tan  a  tan  c  =  — 
cos  z  =  —tan  a  tan  b  =  — 


The  axes  can  be  laid  off  as  in  Fig.  19,  except  that 
while  OA  equals  i,  OM  is  18.  The  drawing  of  a 
structure  in  the  one-third  dimetric  is  made  in  all 
respects  like  the  one-half  dimetric,  except  that  one- 
third  is  measured  along  the  OB  axis  instead  of 
one-half. 

PROBLEM  35.     Show  in  the  (m  :  i  :  m)  dimetric 

that  cos  y=--_. 

PROBLEM  36.  Show  in  the  ( i  :  m  :  i )  dimetric 
that  cos  y  = —  . 


PROBLEM  37.  In  the  system  ( I  :  %  •  i )  find  the 
cosines  of  x,  y,  and  z,  and  draw  the  cross  in  Fig.  13 
in  this  system. 

PROBLEM  38.  In  the  system  ( i  :  %  :  i )  show 
that  cos  y  =  —  2/9,  and  draw  the  monument  in 


38  BACKBONE   OF  PERSPECTIVE. 

Problem  20. 

37.  Trimetric  System.  —  (l:m:n).  In  this  sys- 
tem the  ratios  are  all  different,  but  must  fulfill  the 
conditions  of  formula  (3). 

Find  the  angles  between  the  axes  for  the  (l/z  :  % 
:  i  )  system. 

P2  =  H  -  * 

cos2a  =  ^,  cos2b  =  i,  cos2c  =  f  . 

As  no  cosine  can  be  greater  than  unity,  the  sys- 
tem is  impossible. 

For  the  system  (9  :  10  :  n)  we  have 
2=  2  2 

81  +  100  +  121 
cos2a  =  J||,  cos2b 
sin2a  =  Jf  $,  sin2b 
tan2a  =  iff,  tan2b 
cos  x  ==  —tan  b    tan  c 
cos  y  =  -tan  a    tan  c  =  -VH 
cos  z  =  —tan  a    tan  b  =  - 


Log  COS  X=:^  [log  IO2+log  60  —  log  2OO  —  log  242] 
log  102  =  2.008600  log  200  —  2.301030 

log  60  =  1.778151     log  242  =  2.383815 

3.786751  4.684845 

.-.  Logcosx  =  y2  [3.786751  —4.684845] 

=  J4  [19.101906  —  20]  =9.550953 


Log  cos  y=^2{log  i4O+log  60  —  log  162  —  log  242] 
log  140  =  2.146128  log  162  =  2.209515 

log  60=:  1.778151  log  242  =^2.38381  5 

3.924279  4-593330 


BACKBONE   OF  PERSPECTIVE.  39 

Log  cos  y  =  y2  [3-924279  —  4.593330] 

=  */2  [  I9.330949  —  20]  =  9.6654745 

.-.y=  117°  34  25" 

Log  cos  z=  */<  [log  1  40+  log  1  02  —  log  162  —  log  200] 
log  140  =  2.146128  log  162  =  2.209515 

log  1  02  —  2.008600  log  200  —  2.301030 

4.154728  4.5*0545 

Log  cos  z=  y2  [4.154728  —  4.5!0545] 

=  y2  [19.644183  —  20]  =9.8220915 

•••  2=131°  35'  48" 
Check:    x  +  y  +  z  =  360°. 

38.  Example.  —  Draw  the  perspective  of  a  4" 
cube  that  has  a  hole  i"  square  connecting  each  set 
of  opposite  faces,  the  axis  of  the  hole  coinciding 
with  the  axis  of  the  cube,  in  the  (4:5:6)  system. 

9 


16  +  25  +  36 

cos2a  =  f|,  sin2a  =  4f  ,  tan2a  =  ||. 
cos2b  =  -?4,  sin2b  =  14,  tan2b  =  fj. 
cos2c  =  ,  sin2c  =  -5,  tan2c  =  -. 


cos  x  =  -tan  b  tan  c  =  -Vfiff^  •'•  x  =  l®l°    $'  W 
cos  y  -  -tan  a  tan  c  =  -VfUfj  -'.  Y  ==  108°  12r  36/x 


cos  z  -»  -tan  a  tan  b  -  -VfUfj  •*•  z  =  iSO0  37r  27X/ 
Check  :     x  +  y  +  z  =  360°  0'  0". 

This  system  is  equivalent  to  (4/6:5/6:1). 
Draw  a  vertical  line  for  the  axis  OA  (Fig.  20)  and 
draw  OB  making  angle  AOB  =  108°  12'  30",  and 
angle  AOC=  150°  37'  27".  The  angle  BOC  is  then 
=  101°  9'  57".  Now  as  full  sizes  are  laid  off  on 
OC  and  the  dimension  on  OA  must  be  reduced 


40  BACKBONE   OF  PERSPECTIVE. 

by  multiplying  by  4/6,  or  2/3,  we  can  use  the  4  o 
scale  on  OC  and  the  6  o  scale  on  OA,  as  this  makes 
the  reduction  for  us.  Similarly  we  could  use  the 
5  o  scale  on  OC  and  the  6  o  on  OB.  The  cube  is 
then  easily  drawn. 

PROBLEM  39.  Draw  a  cube  in  the  (3  14  15)  sys- 
tem. 

PROBLEM  40.  Find  the  angles  x,  y,  and  z  in  the 
(5  : 6  17)  system. 

PROBLEM  41.  Find  the  angles  x,  y,  and  z  in  the 
(10  :  ii  :  12)  system. 

PROBLEM  42.  Find  the  angles  x,  y,  and  z  in  the 
(8  :  15  :  17)  system. 


CHAPTER  FOUR. 


Shades  and  Shadows. 

39.  Shade. — An  opaque  body  exposed  to  light 
will  be  illuminated  on  the  side  next  the  light  and 
dark  on  the  opposite  side.  Thus  if  an  opaque 


Figure  21. 

sphere  be  exposed  to  a  source  of  light  P,  a  set  of 
rays  of  light  from  the  source  P  will  be  tangent  to 
the  sphere.  If  the  source  of  light  is  a  point  P  the 
rays  of  light  that  are  tangent  to  the  sphere  will 
form  the  surface  of  a  cone  whose  vertex  is  the 
source  of  light  P  and  which  touches  the  sphere  in 
the  circle  of  contact,  ABC.  The  part  ABCK  of  the 
body,  that  does  not  receive  any  light,  is  in  shade  and 
the  line  of  the  contact  ABC  is  the  curve  of  shade. 
That  part  of  space  from  which  the  source  of  light 
cannot  be  seen  is  called  the  shadow  in  space. 

40.  Shadow. — If  light  is  excluded  from  a  sec- 
ond body  MN,  that  part  of  the  surface  of  the  second 
body  from  which  the  source  of  light  cannot  be  seen 
is  said  to  be  in  shadow,  and  the  dark  part  is  called 
shadow.  The  line  of  contact  of  the  second  body 

41 


42  BACKBONE   OF  PERSPECTIVE. 

with  the  cone  of  rays  tangent  to  the  first  is  called 
the  curve  of  shadow. 

Thus  in  Fig.  21,  P  is  the  source  of  light,  ABCK 
is  an  opaque  sphere  and  MN  is  another  body  that 
intersects  the  shadow  cone  in  the  curve  of  shadow 
DEFG.  The  area  DEFG  is  called  the  shadow  area 
of  the  sphere  on  the  body  MN. 

41.  Drawings   in   Projection. —  The   ordinary 
drawings  of  structures  consist  of  a  side  and  an  end 
view  called  elevations,  and  a  top  view  called  the 
plan.     If  the   structure   is  a  house,   we  ordinarily 
have  the  end  elevation,  side  elevation,  and  a  rear 
elevation. 

In  'addition  to  this  we  have  the  ground  plan  and 
roof  plan  and  these  elevations  and  plans  define  the 
structure  completely,  from  which  we  get  a  concep- 
tion of  the  structure  as  a  whole. 

Projective  drawings  are  referred  to  a  horizontal 
plane  (called  H)  and  a  vertical  plane  (called  V), 
while  their  intersection  is  called  the  ground  line 
(G.  L.).  The  phrases  "horizontal  projection"  and 
"vertical  projection"  will  hereafter  be  abbreviated 
into  hp  and  vp  respectively. 

42.  Shadow  of  Points. — The  shadow  of  a  point 
on  a  plane  is  where  the  ray  of  light  through  the 
point  cuts  the  plane.    If  the  point  is  a  material  point 
it  is  supposed  to  intercept  the  light  and  its  shadow 
will  appear  as  a  dark  spot  on  the  plane. 

43.  Directions  of  the  Rays  of  Light. — The  con- 
ventional direction  of  any  ray  of  light  is  such  that 
the  elevation  and  plan  of  the  ray  make  45°  with 
the  ground  line.     The  ray  of  light  is  supposed  to 
come  over  the  left  shoulder  as  we  face  the  drawing 


BACKBONE   OF  PERSPECTIVE.  43 

in  such  a  way  that  its  projections  make  45°  with 
the  ground  line  (G.  L.). 

44.  Angle  with  H  or  V. — In  Fig.  22,  let  M  be 
a  plane  perpendicular  to  H  and  V  cutting  H  in  LA 
and  V  in  LB. 

PQ  is  a  ray  of  light  whose  hp  is  pQ,  vp  is  p'Q 
and  whose  projection  on  M  is  LR.  The  lines  pQ 


Figure  22. 


and  p'Q  make    45°     with   GL.      .-.  CQp'=45°  = 
CQp.     But  if  angles   CQp   and   CQp'=45°>  then 


Let  CQ  =  a,  then  Cp  =  a 

•*-  PQ 
Tangent 


.•'.  PQp  =  35°  16' 

Now  R  is  the  projection  of  P  on  M,  and  PR  is 
parallel  to  GL  and  perpendicular  to  M.  Hence  LR 
is  the  projection  of  the  ray  PQ  on  M.  Draw  RA 
and  RB  J.  to  H  and  V.  As  GL  is  J-  to  M,  pA  and 
p'B  will  be  parallel  to  GL. 


Cp'  =  LB 
But  CQ  =  Cp  =  Cp' 

.-.  LA  =  LB  or  LA  =  RA 

.-.  The  angle  RLA  is  45° 


44 


BACKBONE    OF  PERSPECTIVE. 


Hence  the  projection  of  the  ray  of  light  on  a  plane 
M  J-  to  H  and  V  makes  an  angle  of  45°  with  lines 
j.  to  H  and  V,  or  with  new  ground  line  LA. 

45.  Shadow  of  Vertical  Rod. — a.  Shadow  all 
on  H.  Let  cd-c'd'  Fig.  23  (a)  be  the  projec- 
tions of  a  vertical  rod.  To  find  the  shadow  of  a 
point  C  on  H,  draw  rays  through  c  and  c',  making 
45°  with  GL.  Find  the  horizontal  trace  at  C'. 


c' 

\ 

c' 

\ 

J\ 

P' 

3  d1 

X 

>     \c 

\       J^ 

V      # 

D' 

\ 

\ 

d 

b    (/ 

. 

(a)        P 

rx7^ 

C 

/ 

Figure  24 

Figure  23. 


The  point  D  is  in  H  and  is  its  own  shadow.  The 
shadow  of  the  rod  will  be  the  heavy  line  D'C'. 

b.  Shadow  on  H  and  V. — Through  cc'  draw 
projections  of  a  ray  making  45°  with  GL  and  find 
the  vertical  trace  at  C'.  Find  the  shadow  P'  of 
some  point  P  on  the  rod.  Join  P'c  and  produce  to 
cut  GL  at  3 ;  then  join  C'3.  The  broken  line  C'3C 
is  the  shadow.  The  part  C'3  on  V  makes  90°  with 
GL,  while  the  part  C3  on  H  makes  45°. 

46.  Horizontal  Lines. — a.  Shadow  on  V. 
Let  cd-c'd'  be  the  projection  of  a  line  perpendicu- 
lar to  V.  Fig.  24  (a).  Draw  rays  through  c  and 


BACKBONE    OF  PERSPECTIVE. 


45 


c'  and  find  shadow  of  C  at  C'.  In  the  same  way 
find  the  shadow  of  D  at  D'.  Connect  C'  and  D', 
giving  the  shadow  CD'. 

b.  Shadow  on  H  and  V.  Fig.  24  (b).  Find 
the  shadow  of  point  C  at  C'  and  D  at  D',  also  find 
the  shadow  of  P  at  P'.  Join  C'P'  to  cut  GL  at  3 
and  then  join  3D'.  The  broken  line  D'3C'  is  the 
shadow  required. 

47.   Shadow  of  Any  Line. — Given  a  line  ab-aV. 

i .  Find  the 
shadow  of  A  at 
A',  and  shadow 
of  B  at  B'.  2. 
Take  any  inter- 
mediate point  P 
on  the  line  AB 


P 

Figure  25. 

Join  A'  and  K'*. 


and  find  its  shad- 
ow at  P'.  Produce 
B'P'  to  cut  GL  at 

K'.    Join  A'  and  K'-.    The  broken  line  A'K'B'  is  the 

shadow  of  the  line  AB. 

48.  Shadow  of  a  Shelf.— Let  a'b'c'f  be  the  ver- 
tical projection  of  the  shelf  and  aebc  the  horizontal 
projection. 

1.  Find  shadow  of  the  line 
cb-c'b'at  C'B'; 

2.  Find  shadow  of  line  ab- 
a'b'  at  A'B'; 

3.  Find  shadow  of  a  line 
cf-c'f  in  c'C'. 

.-.Area     e'A'B'C'c'b'e'     is 
the  shadow  of  the  shelf  on  V  as  shaded  in  Fig.  26. 

49.  Shadow  of  Block  on  H. — Let  abed  be  the 


46  BACKBONE   OF  PERSPECTIVE. 

plan  of  the  top  and  a'b'c'f  be  the  elevation  of  a 
rectangular  block  resting  on  H. 

1.  Find  shadow  of  line 
bc-b'c'  at  B'C' ; 

2.  Find  shadow  of  line 
cd-c'd'  at  CD' ; 

3.  Find  shadow  of  line 
df-d'f  at  DM ; 

.   4.     Find  shadow  of  line 
be-b'e'  at  B'e. 

The      area      bB'C'D'dcb 

(shaded)  is  the  shadow  of 

block  on  H.  Fi*ure  27- 

50.  Shadow  of  a  Circular  Disc  on  H. — Given 
the  projections  ef-e'f  of  the  circular  disc  whose 
plane  is   parallel  to   H    and    perpendicular  to  V. 

The  rays  of  light  that 
touch  the  circle  form  a 
cylindrical  surface  of  rays 
which  is  cut  by  H  parallel 
to  plane  of  circle.  Hence 
the  section  cut  by  H  from 
cylinder  of  rays  is  similar 
to  that  cut  by  the  plane 
of  circle.  Thus  the  shad- 
ow is  a  circle.  To  find  this 
shadow,  find  the  shadow 
of  the  center  of  the  circle 
at  C'and  with  C  as  a  cen- 
Figure  28.  ter  and  radius  equal  to 

that  of  the  circle  draw  the  circle  A'B'D',  which  is 

the  required  shadow. 

51.  Shadow  of  a  Horizontal  Circular  Disc. — 


€ 


BACKBONE   OF  PERSPECTIVE. 


47 


cf  bf 


Given  a  circle  abcd-a'b'c'd'  whose  plane  is  horizon- 
tal and  perpendicular  to  V.  Find  the  shadow  of  a 
series  of  points  as 
a-a',  b-b',  c-c,  d-d/ 
at  A',  B',  C',  D',  etc. 
Through  these 
points  draw  a  curve. 
In  this  case  the 
shadow  falls  on  V 
and  forms  an  ellipse. 

52.  Construct  the 
shadow  of  a  hori-- 
zontal  circular  disc 
where  its  center  is 
equidistant  from  H 
and  V.  Let  abcd-a'b'  d 
c'd'  be  the  projec- 
tions of  the  circular 

d'  c'         o'         a'  b' 


Figure  29. 


the 
the 


disc.  Find 
shadow  of 
two  semi-circu- 
lar halves  abc 
and  adc.  The 
part  of  shadow 
that  falls  on  H 
is  a  circular 
curve,  while  that 
on  V  is  an  ellip- 
tical curve.  First 
find  the  shadow 
of  center  o-o'  at 
O'  and  with  cen- 
ter O'  and  radius  equal  to  oa  draw  semicircle  B'A' 


48  BACKBONE   OF  PERSPECTIVE. 

K'.    Then  construct  elliptical  part  B'C'K'  by  points. 
53.  A  circular  disc  2"  in  diameter  touches  H  at 
a  point  3"  in  front  of  V  and  its  plane  is  perpen- 
dicular to   V.     Construct   its   shadow.     Let  abed- 


Figure  31. 


a'b'c'd'  be  points  on  the  circumference  of  the  cir- 
cular disc.  Find  the  shadows  of  these  points  at 
A'B'CD',  etc.,  and  sketch  a  curve  through  these 
points.  The  shadow  will  be  an  ellipse.  If  we  have 
vp  (b')  of  a  point  B,  the  hp  can  be  found  by  draw- 
ing the  circle  O.  Draw  b'B"  parallel  to  GL  to  cut 
circle  and  diameter  at  B"  and  K.  Lay  off  KB" 


BACKBONE   OF  PERSPECTIVE. 


49 


from  a  on  ac  making  ab  —  KB".     In  the  same  way 
the  projections  of  the  other  points  can  be  found. 

54.  The  plane  of  a  circular  disc  2"  in  diameter 
is  parallel  to  V  and  perpendicular  to  H.  Its  center 
is  3"  from  V  and  i"  from  H.  Construct  its  shadow. 


(See  Fig.  32.)  Locate  points  abcd-a'b'c'd'  and  find 
their  shadow  at  A',  B',  C',  D',  etc.  Draw  the  curve 
through  them,  defining  the  ellipse  of  shadow. 

55.  Shadow  of  a  Chimney. —  In  Fig.  33  let 
epqf  be  the  plan  of  a  hip  roof  and  p'e'f's'  the  eleva- 
tion, while  abc  and  a'b'c'  represent  the  plan  and 
elevation  of  a  chimney.  The  line  ab-a'b'  is  perpen- 
dicular to  V  and  its  shadow  on  H  will  be  in  the  line 
ef,  which  is  found  by  drawing  a'e'  through  a'  at 


50  BACKBONE   OF  PERSPECTIVE. 

45°  with  GL.  Produce  e'e  to  cut  the  eaves  of  roof 
at  e  and  f.  If  the  line  AB  were  indefinite  in  extent, 
its  shadow  on  the  roof  would  pass  through  e  and  f. 
The  shadow  of  AB  will  cut  the  comb  of  the  roof  at 
o-o'  where  a'e'  cuts  the  comb  s't'.  Then  the  shadow 


Figure  33. 


of  AB  will  lie  in  the  lines  oe  and  fo.  Through  a 
and  b  draw  rays,  ag  and  bh,  making  45°  with  GL 
cutting  oe  at  g  and  fo  at  h.  The  broken  line  goh 
will  be  the  shadow  of  ab-a'b'  on  the  roof.  The 
shadow  of  the  line  bc-b'c'  will  be  parallel  to  GL  and 
hence  will  be  mh;  while  the  shadow  of  the  vertical 
corner  ak-a'k'  will  be  ag ;  and  that  at  C  will  be  cm. 
The  shaded  area  agohmcba  will  be  the  plan  of  the 


BACKBONE   OF  PERSPECTIVE.  51 

shadow,  and  k'g'o's't'  will  be  the  elevation  of  the 
shadow. 

56.   Shadow  of  Cap  on  Cylinder. — Given  a  ver- 
tical cylinder  2"  diameter  and  a  cap  3^' ',  required 


Figure  34. 

to  find  shadow  of  cap  on  the  cylinder.  See  Fig.  34. 
Let  abc-a'b'c'  be  the  projections  of  the  cap  and 
efg-e'f'g'  the  projections  of  the  cylinder.  Draw 
diameter  eh  parallel  to  GL,  and  from  e  draw  ray  ea 
at  45°  with  GL.  Mark  a'  vertically  above  a.  Draw 
a' A'  at  45°  with  GL,  cutting  a  vertical  through  e  at 
A'.  Draw  any  intermediate  ray  through  bb',  cutting 
surface  of  cylinder  at  f.  Locate  b'  vertically  above 
b,  and  draw  b'B'  at  45°  with  GL  to  cut  vertical 
through  f  at  B'.  The  extreme  point  of  shadow 


52  BACKBONE    Ob    PERSPECTIVE. 

curve  will  be  above  g  where  Og  makes  45°  with 
GL.  Draw  gc  perpendicular  to  Og,  cutting  rim  of 
cap  at  c.  Find  c'  vertically  above  c  and  draw  c'C 
at  45°  with  GL  to  cut  vertical  through  g  at  C'. 
The  line  A'B'C'  is  the  shadow  line.  The  rest  of  the 
visible  part  of  the  cylinder  to  right  of  vertical 
through  g  is  in  shade  as  indicated  by  area  h'g'.  In 
the  same  way  the  part  of  cap  to  right  of  vertical 
through  k  is  in  shade. 

57.    Shadow  of  a  Straight  Line  with  the  Pro- 
jections   Perpendicular    to    GL. —  Let  ab-a'b'  be 


the  projections  of  a  straight  line  where  ab  and 
a'b'  are  perpendicular  to  GL,  in  Fig.  35.  Find  the 
shadow  of  A  at  A'  and  B  at  B'.  If  the  line  AB  is 
revolved  around  its  vertical  projection  into  the  ver- 
tical plane,  it  will  appear  at  AB  and  its  horizontal 
trace  and  vertical  traces  will  be  D  and  E  respectively. 
If  these  points  D  and  E  are  taken  back  to  their  true 


BACKBONE   OF  PERSPECTIVE. 


53 


positions  they  will  appear  at  D'  and  E'  and  each 
point  will  be  its  own.  shadow.  Join  D'A'  and  pro- 
duce to  cut  GL  at  C  and  then  join  CB'.  The  broken 
line  A'CB'  will  be  the  required  shadow.  The  shad- 
ows of  all  lines  parallel  to  AB  will  be  parallel  to 
A'C  and  B'C. 

58.    Shadow  of  Wall  on  Steps.— Let  i,  2,  3,  4, 
5,  6,  7  be  the  end  view  of  a  series  of  steps  and  ABC 


i       !     ! 
— j— -B-H-- i-— +-H 

/  I 


that  of  a  wing-wall,  in  Fig.  36.  The  vertical  and 
horizontal  projections  are  shown  above  and  below 
GL.  Now  the  shadow  of  a  line  AB  is  found  as  in 
Art.  57  in  the  broken  line  E'KD'.  That  part  D'c 
is  on  H  and  at  point  c  the  shadow  strikes  the  verti- 


54  BACKBONE   OF  PERSPECTIVE. 

cal  face  of  the  first  step.  The  shadow  on  the 
vertical  face  of  the  first  step  will  pass  through  c' 
and  be  parallel  to  KE' ;  i.  e.,  c'e'  will  be  this  shadow. 
Through  e,  the  horizontal  projection  of  e'  draw  ef 
parallel  to  D'K.  The  shadows  on  the  other  steps 
will  be  parallel  to  c'e'  and  ef  respectively.  The 
shaded  areas  represent  the  shadows  on  the  vertical 
and  horizontal  planes  of  the  steps. 

59.  Shadow  on  a  Semicylinder. — Let  abc-a'b'c' 
represent  the  horizontal  and  vertical  projections  of 

a  semicylindrical  sur- 
face. (See  Fig.  37.) 
Required  to  find  the 
shadow  of  the  rim  on 
the  cylindrical  surface. 
Through  a  draw  ad  at 
45°  with  GL  to  cut  the 
cylindrical  surface  at  d 
and  through  a'  draw 
a'A'  at  45°  with  GL  to 
cut  vertical  through  d 
at  A'.  To  find  the 
shadow  of  any  inter- 
mediate point  C,  through  c  draw  C3  at  45°  with  GL 
to  cut  surface  of  cylinder  at  3.  Through  c'  draw 
c'C'  at  45°  with  GL  to  cut  vertical  through  3  at  C'. 
If  Ob  makes  45°  with  GL,  b'  is  a  point  in  shadow 
line.  Thus  the  shadow  of  rim  is  A'C'b'.  The  area 
a'b'A'd'G  is  the  required  shadow. 

60.  Shadow  of  Half-Cylinder  on  H.  —  Given 
abed  and  a'  c',  in  Fig.  38,  the  horizontal  and  ver- 
tical projections  of  a  half-cylinder  whose  axis  is 
parallel  to   GL.     It   is   required  to   construct  the 


BACKBONE   OF  PERSPECTIVE. 


55 


shadow  cast  by  the  cylinder  on  its  curved  surface 
and  on  H.  The  shadow  of  the  semi-circle  ab-aV 
will  be  a  semi-ellipse  A'OB'  which  is  found  as  in 
Art.  53.  That  of  cd-c'd'  is  similarly  found,  but 


only  that  part  C'E'  will  be  seen.  The  shadow  of 
the  left  end  on  the  interior  surface  of  the  cylinder 
will  be  found  as  in  Art.  59.  The  shaded  area  D' 
Oad  is  on  the  cylinder  while  A'gOB'C'E'b  is  on  H. 
61.  Shadow  of  Cone. — Let  Fig.  39  represent  a 
right  cone  whose  axis  is  vertical  with  base  resting 
on  the  horizontal  plane.  Find  the  shadow  of  the 
vertex  at  A'  and  from  A'  draw  A'b  and  A'c  tan- 
gent to  the  circle  bee.  The  area  A'cebA'  will  be 
the  shadow  that  the  cone  casts  on  H.  The  area 
cebac  is  the  horizontal  projection  of  the  shade  cast 
by  the  cone  on  its  own  surface,  while  a'b'f  is  the 


56  BACKBONE   OF  PERSPECTIVE. 

vertical  projection  of  that  part  of  the  shade  that  is 
visible.  Fig.  40  shows  how  the  shadow  of  the  cone 
falls  when  the  shadow  of  the  vertex  falls  on  the 
vertical  plane.  Draw  the  rays  of  light  through  the 
vertex  at  45°  with  GL.  The  horizontal  trace  of  the 
ray  will  be  at  A"  and  its  vertical  trace  at  A'.  From 


Figure  39.  Figure  40. 

A"  draw  lines  A"b  and  A"c  tangent  to  the  circle. 
Join  the  points  where  A"b  and  A"c  cut  GL  to  the 
vertical  trace  A'.  The  shaded  area  A'ceb  will  be 
the  shadow  of  the  cone  on  the  horizontal  and  ver- 
tical planes.  The  areas  ceba  and  a'b'f  will  be  the 
horizontal  and  vertical  projections  of  the  shade 
the  cone  casts  on  its  own  surface. 


OF  **  ***** 


THIS  BOOK  ON  THE  DATE  *  n"**  T°  '— 
WILL  'NCREASETO  50  CEN?S  n  ™E  PENALTY 
DAY  AND  TO  $,  OO  ON  TM*  THE  FOU*TH 
OVERDUE.  '  THE  SEVENTH  DAY 


21-loom-8,'34 


/C/f 


VD       !  r-o 

YB    loo 


